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CED objectives

Compound Direct Current Circuits

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletCompound Lab · a resistor in series with a parallel block and a switch that adds a branch, with two wrong reductions printed in red

Compound circuits are solved by collapsing them. Two elements are in parallel when they share both nodes, so they have one voltage and split the current; they are in series when they share one node that connects nothing else, so they have one current and split the voltage. Series resistances add, parallel conductances add, and a parallel equivalent always comes out smaller than every branch. Reduce a ladder from the far end inward, relabeling after each collapse, then work back out to recover each branch's current and drop.

Four errors dominate. Classifying elements from how the diagram is drawn instead of from the node list. Adding parallel resistances directly, or stopping at $1/R_{\text{eq}}$ and reporting it as $R_{\text{eq}}$. Collapsing two resistors as series when a third branch taps the node between them, which poisons every later step of the reduction. And splitting a source voltage evenly among unequal series resistors, when the drops go as $IR_i$ and are therefore in the ratio of the resistances. Behind several of them sits the intuition that more components must mean more resistance, when a parallel branch is another road and lowers $R_{\text{eq}}$ while raising the total current.

parallel combination lands BELOW the smallest branch 6 Ω branch 1 3 Ω branch 2 smallest branch 2 Ω 6 in parallel with 3 9 Ω 6 + 3, the wrong move 1/R = 1/6 + 1/3 = 1/2, then FLIP: R = 2 Ω. Stopping at 1/2 reports 0.5 Ω.
Two branches at 6 and 3 ohms combine to 2 ohms. Any parallel answer at or above the smallest branch is wrong before the algebra is checked, and 0.5 ohms is the same arithmetic with the final flip left off.
12 V across 2 Ω and 10 Ω in series, one current of 1 A 2 V 10 V V = IR, so the drops are in the ratio 1 : 5, exactly the resistance ratio 6 V 6 V the even split: correct only when the resistors are equal
Series elements share a current, so the source voltage divides in proportion to resistance. An even split is the equal-resistor special case, not the rule.

The work

3 ways in · any order
Lesson
Compound Direct Current Circuits

Builds the node list first, then classifies series and parallel from it, drills the reciprocal combination against its size bound, and orders a ladder reduction around the branch points that forbid a series collapse.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: classifying elements from the drawing instead of the node list, adding parallel resistances directly or forgetting the final flip, collapsing across a junction, expecting a parallel branch to raise the total resistance, and splitting series voltage evenly. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions